Xrfgtjtk

I gave this problem a try and I feel like I am missing something and I wanted to get some help on what I might be missing.

Suppose $Omega$ is a countable space and $p:Omegato[0,1]$ is such that $sum_{omegainOmega}p(omega)=1$.
For $AsubsetOmega$ let $P(A)=sum_{omegain A}p(omega)$ with $P(varnothing)=0$. Prove that if $A_n$, $ninmathbb{N}$, are mutually disjoint, i.e. $A_ncap A_m=varnothing$ for $mne n$, then
$P(cup_n A_n)=sum_n P(A_n)$

My try:

To show this, notice that $P(cup_n A_n)=$ (sum of the measures of the $A_n$) - (sum of the measures of the intersections of the $A_n$). Since the $A_n$ are mutually disjoint, we know that the sum of the intersections is 0. So, it follows that $P(cup_n A_n)=sum_n P(A_n)$.

This is more a question about summation rather than probability.

We need to show that $sum_{omega in A} p(omega) = sum_n sum_{omega in A_n} p(omega)$, where $A= cup_n A_n$ and the $A_n$ are pairwise disjoint.

Note that all the index sets are subsets of $Omega$ which is countable and $p$
is non negative.

It is not hard to show that
$sum_{omega in B} p(omega) = sup_{I subset B, text{ finite}} sum_{omega in I} p(omega)$, and hence if $B subset C$ then
$sum_{omega in B} p(omega) le sum_{omega in C} p(omega)$.

Suppose $I subset A$ is finite, then $I cap A_n$ is finite and we have
$sum_{omega in I} p(omega) le sum_n sum_{omega in I cap A_n} p(omega) le sum_n sum_{omega in A_n} p(omega)$ and hence
$sum_{omega in A} p(omega) le sum_n sum_{omega in A_n} p(omega)$.

Now let $epsilon>0$ and pick $N$ such that $sum_{n le N} sum_{omega in A_n} p(omega) > sum_n sum_{omega in A_n} p(omega) - {1 over 2} epsilon$

Now choose finite $I_n subset A_n$ such that
$sum_{omega in I_n} p(omega) > sum_{omega in A_n} p(omega) - {1 over 2^{n+1}} epsilon$.

Then
begin{eqnarray}
sum_{omega in A} p(omega) &ge& sum_{n le N} sum_{omega in I_n} p(omega) \
&ge& sum_{n le N} (sum_{omega in A_n} p(omega) -{1 over 2^{n+1}} epsilon) \
&ge& sum_{n le N} sum_{omega in A_n} p(omega) -{1 over 2} epsilon \
&ge& sum_n sum_{omega in A_n} p(omega) - epsilon
end{eqnarray}
from which we get the other direction.

EDIT I've realized that this proof is circular since I assumed at the beginning of the proof that $P$ is a measure. Nevertheless, I feel like leaving it here as an example of "math made difficult", in case anyone might find it amusing.

Let $A:=bigcup_n A_n$. We shall consider the characteristic functions
$chi_{A_n}$ and recall that $P$ is the measure whose density function is $p(omega)$.
Your question can be interpreted as proving that
$$begin{align}
P(A) &= sum_{n=1}^infty int_Omega chi_{A_n},dP \
&=lim_{ntoinfty} sum_{i=1}^n int_Omega chi_{A_i},dP \
&= lim_{ntoinfty} int_Omega f_{n},dP
end{align}$$
where $f_n :=sum_{i=1}^n chi_{A_i}$.

By mutual disjoint-ness of $A_i$, we can see that $f_n=chi_{cup_{i=1}^n A_i}$ so $f_nle 1$. It is also easy to verify that $f_ntochi_A$ pointwise, hence we can apply the Dominated Convergence Theorem to conclude that
$$
lim_{ntoinfty} int_Omega f_{n},dP = int_Omega chi_A dP = P(A)
$$
which proves the assertion.

Alternatively, if you want a more "elementary" method then I think you can fix an arbitrary enumeration of $A$ and try to invoke the Riemann rearrangement theorem. I haven't work out the full details but it should be doable.